19
Q:

In an election only two candidates contested 20% of the voters did not vote and 120 votes were declared as invalid. The winner got 200 votes more than his opponent  thus he secured 41% votes of the total voters on the voter list. Percentage votes of the defeated candidate out of the total votes casted is:

A) 47.5% B) 41%
C) 38% D) 45%

Answer:   D) 45%

Explanation:

Let there be x voters and k votes goes to loser then 

      0.8x - 120 = k + (k + 200)

         k+200 = 0.41x

        \inline \Rightarrow  k = 1440

and  (k + 200) =1640

Therefore  \inline \frac{1440}{3200}\times 100 = 45%

Q:

If three eighth of a number is 141. What will be the approximately value of 32.08% of this number ?

A) 101 B) 112
C) 104 D) 120
 
Answer & Explanation Answer: D) 120

Explanation:

K x 3/8 = 141 => K = 376

376 x 32.08/100 = 120

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0 39
Q:

If L exceeds K by 30%, then K is less than L by ?

A) 14 2/5 % B) 23 1/3 %
C) 20 % D) 16 %
 
Answer & Explanation Answer: B) 23 1/3 %

Explanation:

K = 100 L = 130
130------ 30
100------- ? => 23 1/13 %

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2 103
Q:

A user type a content on 20 sheets, each sheet consist of 55 lines, and each line can have 66 characters. The content is again retyped on another set of sheets, these new sheets have 65 lines and each line can take 70 characters. Calculate the reduction in sheets ?

A) 2 or 60% B) 4 or 20%
C) 6 or 10% D) 8 or 5%
 
Answer & Explanation Answer: B) 4 or 20%

Explanation:

20 x 55 x 66 = L x 65 x 70
Here L = 72,600/4550 = 15.95 =~ 16
Reduction is 20-16 = 4

which is 20% reduction.

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2 134
Q:

Kalyan purchased a Maruti van for Rs. 1,96,000 and the rate of depreciation is 14(2/7) % per annum. Find the value of the van after two years ?

A) Rs. 1,24,000 B) Rs. 1,44,000
C) Rs. 1,32,000 D) Rs. 1,62,000
 
Answer & Explanation Answer: B) Rs. 1,44,000

Explanation:

Value of Maruti Van, V1 = Rs. 1,96,000
Rate of depreciation, r = 14(2/7)% = 100/7%; Time, t = 2 years
Let V2 is the value after depreciation.
V2 =
V2 =
V2 = 196000(6/7)(6/7)
V2 = (196000x36)/49 = Rs. 1,44,000.

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3 133
Q:

A duster is of dimension 5x5x7 all in cms,  is sinked into a cylindrical vessel of radius 5 cm and height 10 cm. After insertion the level of the water is inccresed by 2.14 cm. Then by what percentage the part of the duster is above the layer ?

A) 96 % B) 4 %
C) 97 % D) 3 %
 
Answer & Explanation Answer: B) 4 %

Explanation:

Imension of the duster is assumed as 5x5x7 cm3

Volume of the duster below the water layer
= volume of water increased
= πxrxrxh = 3.14×52×2.14 ≈ 168 cm3

Total volume of the duster = 5×5×7 = 175 cm3

Percentage of the duster part below the layer = 168×100/175 = 96%

Percentage of the duster part above the layer = 4%

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2 149